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Consider the infinite geometric series (∞/Σ/n=1) * -4(1/3)^{n-1}.

AWrite the first 4 terms of the series.
BDoes the series converge or diverge?
CIf the series has a sum, find the sum.

Verified Answer :

LammettHash

The first four terms of the series probably refer to the first four partial sums.

\(\displaystyle\sum_{n=1}^1-4\left(\frac13\right)^{n-1}=-4\left(\frac13\right)^{1-1}=-4\)
\(\displaystyle\sum_{n=1}^2-4\left(\frac13\right)^{n-1}=-\frac{16}3\)
\(\displaystyle\sum_{n=1}^3-4\left(\frac13\right)^{n-1}=-\frac{52}9\)
\(\displaystyle\sum_{n=1}^4-4\left(\frac13\right)^{n-1}=-\frac{160}{27}\)

which you can compute either by adding one term at a time, or using the well-known formula,

\(\displaystyle\sum_{n=1}^kar^{n-1}=a\frac{1-r^{k+1}}{1-r}\)

(I can provide a link to a derivation I gave in a nearly identical question in the comments)

The series converges as \(k\to\infty\) if and only if \(|r|<1\), which is certainly the case here, since \(-1<\dfrac13<1\).

Extrapolating from the formula above, the sum of the convergent series is

\(\displaystyle\sum_{n=1}^\infty ar^{n-1}=\frac a{1-r}\)

so the sum for this series is \(\dfrac{-4}{1-\frac13}=-6\).

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